1500 LeetCode Problems
Optimizing 'Adjacent Increasing Subarrays Detection II' with Pre-computation and Caching
Solving algorithmic problems efficiently often hinges on smart strategies like pre-computation and caching. The LeetCode problem “Adjacent Increasing Subarrays Detection II” (https://leetcode.com/problems/adjacent-increasing-subarrays-detection-ii/) serves as a perfect example of how these techniques can be applied to achieve a linear-time solution. This post explores how to implement these strategies to solve the problem effectively.
Understanding the Problem
The challenge is to find the maximum length of a subarray that increases continuously, under specific conditions. The naive approach might involve nested loops, resulting in higher time complexity. The goal, however, is to process the array in linear time, avoiding unnecessary computations.
Strategy Overview
The key to optimizing the solution lies in:
- Pre-computing the lengths of all increasing subarrays.
- Caching these lengths to avoid redundant calculations during the final evaluation.
Step-by-Step Solution
1. Identifying Increasing Subarrays
Create a helper function that scans the array from a given index to find the longest increasing subarray starting from that point.
private int FindLongestIncreasingSubarraysIndex(IList<int> nums, int index)
{
int retVal = index;
for (int i = index + 1; i < nums.Count; i++)
{
if (nums[i] <= nums[i - 1]) break;
retVal = i;
}
return retVal;
}
This function increments the `retVal` as long as the sequence continues to increase, effectively marking the end index of the current increasing subarray.
2. Pre-computing Subarray Lengths
Iterate through the array once to compute and store the lengths of all increasing subarrays.
List<int> indexes = new List<int>();
int index = 0;
while (index < nums.Count)
{
int nextIndex = FindLongestIncreasingSubarraysIndex(nums, index);
indexes.Add(nextIndex - index + 1);
index = nextIndex + 1;
}
By storing these lengths in the `indexes` list, the algorithm caches valuable information for the next step.
3. Calculating the Maximum Length
Use the pre-computed lengths to determine the maximum length according to the problem's conditions.
int retVal = 0;
if (indexes.Count == 1)
{
retVal = indexes[0] / 2;
}
else
{
for (int i = 1; i < indexes.Count; i++)
{
if (indexes[i] < indexes[i - 1])
{
int candidate = Math.Max(indexes[i - 1] / 2, indexes[i]);
retVal = Math.Max(retVal, candidate);
}
else
{
retVal = Math.Max(retVal, indexes[i - 1]);
}
}
retVal = Math.Max(retVal, indexes[indexes.Count - 1] / 2);
}
return retVal;
This code compares adjacent increasing subarrays and updates the `retVal` based on the specified rules, ensuring that all potential candidates for the maximum length are considered.
Full Solution Code
Combining all the steps, the complete solution is as follows:
public class Solution {
public int MaxIncreasingSubarrays(IList<int> nums)
{
List<int> indexes = new List<int>();
// O(n) pre-computation of increasing subarray lengths
int index = 0;
while (index < nums.Count)
{
int nextIndex = FindLongestIncreasingSubarraysIndex(nums, index);
indexes.Add(nextIndex - index + 1);
index = nextIndex + 1;
}
// O(n) calculation of the maximum length
int retVal = 0;
if (indexes.Count == 1)
{
retVal = indexes[0] / 2;
}
else
{
for (int i = 1; i < indexes.Count; i++)
{
if (indexes[i] < indexes[i - 1])
{
int candidate = Math.Max(indexes[i - 1] / 2, indexes[i]);
retVal = Math.Max(retVal, candidate);
}
else
{
retVal = Math.Max(retVal, indexes[i - 1]);
}
}
retVal = Math.Max(retVal, indexes[indexes.Count - 1] / 2);
}
return retVal;
}
private int FindLongestIncreasingSubarraysIndex(IList<int> nums, int index)
{
int retVal = index;
for (int i = index + 1; i < nums.Count; i++)
{
if (nums[i] <= nums[i - 1]) break;
retVal = i;
}
return retVal;
}
}
Why This Approach Works
- Linear Time Complexity: Both the pre-computation and the final calculation run in O(n) time, ensuring the overall solution is efficient.
- Avoiding Redundancy: By caching the lengths of increasing subarrays, the algorithm eliminates the need for repeated calculations.
- Simplicity and Clarity: The code is straightforward, making it easier to understand and maintain.
Conclusion
Utilizing pre-computation and caching can significantly optimize solutions for array-based problems. This approach not only improves performance but also enhances code readability. Applying these techniques to the "Adjacent Increasing Subarrays Detection II" problem demonstrates how thoughtful algorithm design leads to efficient and elegant solutions.
By focusing on strategic pre-computation and effective use of caching, complex problems can often be transformed into manageable tasks with optimal performance. This methodology is a valuable addition to any programmer's toolkit.
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